Question 425592: 1. The graduate selection committee wants to select the top 10% of applicants. On a standardized test with a mean of 500 and a standard deviation of 100, what would be the cutoff score for selecting the top 10% of applicants, assuming that the standardized test is normally distributed?
This is my answer:
First, the z value had to be calculated. Assuming the number would be positive, the z score would be right tail
The z score formula is Z= (X-M)/SD
Z=(90-500)/100
Z=-410/100=
Z=-4.10
A z score of -4.10 gets a p value of <0.01
Did I calculate this correctly? If not please explain. I was not certain if I should use 90, since the question is looking at the top 10%, or 10 since it relates to 10%. Please, help me calculate this answer.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. The graduate selection committee wants to select the top 10% of applicants. On a standardized test with a mean of 500 and a standard deviation of 100, what would be the cutoff score for selecting the top 10% of applicants, assuming that the standardized test is normally distributed?
This is my answer:
First, the z value had to be calculated. Assuming the number would be positive, the z score would be right tail
The z score formula is Z= (X-M)/SD
Z=(90-500)/100
Z=-410/100=
Z=-4.10
A z score of -4.10 gets a p value of <0.01
Did I calculate this correctly? If not please explain. I was not certain if I should use 90, since the question is looking at the top 10%, or 10 since it relates to 10%. Please, help me calculate this answer.
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Yes, you are greatly confused.
If you want the top 10% find the z-value with a right tail of 10%.
Using a TI-84 that would be invNorm(0.9) = 1.2816
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Now, since z = (x-u)/sigma,
x = zs+u
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So the score you want is x = 1.2816*100 + 500 = 128.16+500 = 628.16
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Cheers,
Stan H.
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